<?xml version="1.0" encoding="UTF-8"?><rss version="2.0"
	xmlns:content="http://purl.org/rss/1.0/modules/content/"
	xmlns:dc="http://purl.org/dc/elements/1.1/"
	xmlns:atom="http://www.w3.org/2005/Atom"
	xmlns:sy="http://purl.org/rss/1.0/modules/syndication/"
		>
<channel>
	<title>Comments on: The curve of an Instalanche</title>
	<atom:link href="http://neoneocon.com/2008/12/01/the-curve-of-an-instalanche/feed/" rel="self" type="application/rss+xml" />
	<link>http://neoneocon.com/2008/12/01/the-curve-of-an-instalanche/</link>
	<description></description>
	<lastBuildDate>Thu, 24 May 2012 07:05:20 +0000</lastBuildDate>
	<sy:updatePeriod>hourly</sy:updatePeriod>
	<sy:updateFrequency>1</sy:updateFrequency>
	<generator>http://wordpress.org/?v=3.1.3</generator>
	<item>
		<title>By: Alex</title>
		<link>http://neoneocon.com/2008/12/01/the-curve-of-an-instalanche/#comment-95319</link>
		<dc:creator>Alex</dc:creator>
		<pubDate>Fri, 05 Dec 2008 20:38:37 +0000</pubDate>
		<guid isPermaLink="false">http://neoneocon.com/2008/12/01/the-curve-of-an-instalanche/#comment-95319</guid>
		<description>I&#039;ve got an explanation for the whole e^x=Σ (x^n)/n! thing.  Bear with me.

The function e^x is the only possible function that is always equal to its own derivative.  Its value (for any given x) is always the same as its slope (at that x).  Any other function that is equal to its own derivative must be equivalent to e^x.

If you take the derivative of the infinite series

Σ (x^n)/n! = 1 + x/1! + x^2/2! + x^3/3! + ...

each term in the series &quot;shifts down&quot; one.  The 1 disappears, the x becomes a 1, the x^2/2! becomes x, the x^3/3! becomes x^2/2!, and so on. Because the series is infinite, this means that the derivative is of the series is equal to the series itself.  Since it is its own derivative, it must be equal to e^x.</description>
		<content:encoded><![CDATA[<p>I&#8217;ve got an explanation for the whole e^x=Σ (x^n)/n! thing.  Bear with me.</p>
<p>The function e^x is the only possible function that is always equal to its own derivative.  Its value (for any given x) is always the same as its slope (at that x).  Any other function that is equal to its own derivative must be equivalent to e^x.</p>
<p>If you take the derivative of the infinite series</p>
<p>Σ (x^n)/n! = 1 + x/1! + x^2/2! + x^3/3! + &#8230;</p>
<p>each term in the series &#8220;shifts down&#8221; one.  The 1 disappears, the x becomes a 1, the x^2/2! becomes x, the x^3/3! becomes x^2/2!, and so on. Because the series is infinite, this means that the derivative is of the series is equal to the series itself.  Since it is its own derivative, it must be equal to e^x.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: &#187; Thursday Night Links NineCents: The Blog That Makes No Cents</title>
		<link>http://neoneocon.com/2008/12/01/the-curve-of-an-instalanche/#comment-95269</link>
		<dc:creator>&#187; Thursday Night Links NineCents: The Blog That Makes No Cents</dc:creator>
		<pubDate>Fri, 05 Dec 2008 05:25:22 +0000</pubDate>
		<guid isPermaLink="false">http://neoneocon.com/2008/12/01/the-curve-of-an-instalanche/#comment-95269</guid>
		<description>[...] Neo-Neocon examines the instalanche, and the commenters tackle it with math. [...]</description>
		<content:encoded><![CDATA[<p>[...] Neo-Neocon examines the instalanche, and the commenters tackle it with math. [...]</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Ymarsakar</title>
		<link>http://neoneocon.com/2008/12/01/the-curve-of-an-instalanche/#comment-94968</link>
		<dc:creator>Ymarsakar</dc:creator>
		<pubDate>Wed, 03 Dec 2008 01:31:11 +0000</pubDate>
		<guid isPermaLink="false">http://neoneocon.com/2008/12/01/the-curve-of-an-instalanche/#comment-94968</guid>
		<description>&lt;b&gt;Jamie, the mystery to me is the d/dx (e^x) = e^x. Spooky. You can diffentiate as many times as you like, but each term lopped off at the beginning is replaced by the one following, no matter how fast you differentiate (/g).&lt;/b&gt;

There&#039;s a proof somewhere floating around. Since e is a constant and x is the variable, it is like taking a derivative of A to the x.

Some weird thing with natural logs, however, means that e to the x is its own derivative. 

&lt;b&gt;To me an even greater mystery is that e^x should be represented by

Σ (x^n)/n!&lt;/b&gt;

Maybe you could blame Taylor for that one.</description>
		<content:encoded><![CDATA[<p><b>Jamie, the mystery to me is the d/dx (e^x) = e^x. Spooky. You can diffentiate as many times as you like, but each term lopped off at the beginning is replaced by the one following, no matter how fast you differentiate (/g).</b></p>
<p>There&#8217;s a proof somewhere floating around. Since e is a constant and x is the variable, it is like taking a derivative of A to the x.</p>
<p>Some weird thing with natural logs, however, means that e to the x is its own derivative. </p>
<p><b>To me an even greater mystery is that e^x should be represented by</p>
<p>Σ (x^n)/n!</b></p>
<p>Maybe you could blame Taylor for that one.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Harry</title>
		<link>http://neoneocon.com/2008/12/01/the-curve-of-an-instalanche/#comment-94904</link>
		<dc:creator>Harry</dc:creator>
		<pubDate>Tue, 02 Dec 2008 19:09:08 +0000</pubDate>
		<guid isPermaLink="false">http://neoneocon.com/2008/12/01/the-curve-of-an-instalanche/#comment-94904</guid>
		<description>I captured an Instalanche sitemeter grab as well, complete with the quantitative hit data. You can see it &lt;a href=&quot;http://flickr.com/photos/hforbes/3077286331/sizes/o/&quot; rel=&quot;nofollow&quot;&gt; here. &lt;/a&gt;</description>
		<content:encoded><![CDATA[<p>I captured an Instalanche sitemeter grab as well, complete with the quantitative hit data. You can see it <a href="http://flickr.com/photos/hforbes/3077286331/sizes/o/" rel="nofollow"> here. </a></p>
]]></content:encoded>
	</item>
	<item>
		<title>By: njcommuter</title>
		<link>http://neoneocon.com/2008/12/01/the-curve-of-an-instalanche/#comment-94874</link>
		<dc:creator>njcommuter</dc:creator>
		<pubDate>Tue, 02 Dec 2008 17:22:16 +0000</pubDate>
		<guid isPermaLink="false">http://neoneocon.com/2008/12/01/the-curve-of-an-instalanche/#comment-94874</guid>
		<description>Neo, that a couple of your regulars can toss off the basics of the exponential function and its derivatives speaks well of your readership and, I expect, of your blog.

I love the way the mathematicians define &lt;i&gt;e&lt;/i&gt;: the limit as &lt;i&gt;n&lt;/i&gt; goes to zero of the &lt;i&gt;n&lt;/i&gt;th root of ( 1 + &lt;i&gt;n&lt;/i&gt;).  Identical to &lt;b&gt;Jamie Irons&lt;/b&gt;&#039;s statement, but good for making a jaw drop.  (Real mathematicians may be able to explain a difference, unknown to mere mortals, between the two formulations.  The invitation is extended.)</description>
		<content:encoded><![CDATA[<p>Neo, that a couple of your regulars can toss off the basics of the exponential function and its derivatives speaks well of your readership and, I expect, of your blog.</p>
<p>I love the way the mathematicians define <i>e</i>: the limit as <i>n</i> goes to zero of the <i>n</i>th root of ( 1 + <i>n</i>).  Identical to <b>Jamie Irons</b>&#8216;s statement, but good for making a jaw drop.  (Real mathematicians may be able to explain a difference, unknown to mere mortals, between the two formulations.  The invitation is extended.)</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Alex</title>
		<link>http://neoneocon.com/2008/12/01/the-curve-of-an-instalanche/#comment-94865</link>
		<dc:creator>Alex</dc:creator>
		<pubDate>Tue, 02 Dec 2008 14:56:02 +0000</pubDate>
		<guid isPermaLink="false">http://neoneocon.com/2008/12/01/the-curve-of-an-instalanche/#comment-94865</guid>
		<description>Wow.  You must have been a very active 10 year old.</description>
		<content:encoded><![CDATA[<p>Wow.  You must have been a very active 10 year old.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: SteveH</title>
		<link>http://neoneocon.com/2008/12/01/the-curve-of-an-instalanche/#comment-94854</link>
		<dc:creator>SteveH</dc:creator>
		<pubDate>Tue, 02 Dec 2008 11:50:05 +0000</pubDate>
		<guid isPermaLink="false">http://neoneocon.com/2008/12/01/the-curve-of-an-instalanche/#comment-94854</guid>
		<description>Hey that looks just like a bar graph of my sex life.</description>
		<content:encoded><![CDATA[<p>Hey that looks just like a bar graph of my sex life.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: br549</title>
		<link>http://neoneocon.com/2008/12/01/the-curve-of-an-instalanche/#comment-94849</link>
		<dc:creator>br549</dc:creator>
		<pubDate>Tue, 02 Dec 2008 09:51:16 +0000</pubDate>
		<guid isPermaLink="false">http://neoneocon.com/2008/12/01/the-curve-of-an-instalanche/#comment-94849</guid>
		<description>Oh, knock it off.</description>
		<content:encoded><![CDATA[<p>Oh, knock it off.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Alex</title>
		<link>http://neoneocon.com/2008/12/01/the-curve-of-an-instalanche/#comment-94836</link>
		<dc:creator>Alex</dc:creator>
		<pubDate>Tue, 02 Dec 2008 05:35:32 +0000</pubDate>
		<guid isPermaLink="false">http://neoneocon.com/2008/12/01/the-curve-of-an-instalanche/#comment-94836</guid>
		<description>Here are two simplified models of the world that would lead to exponential decay:

1.  Instapundit readers arrive at the site randomly, and once there they read all the posts they have not read yet. As time goes on more and more of the readers are repeat visitors, and they don&#039;t re-read.  To take one simple example, suppose there were 100 readers and each had a 50% chance of visiting Instapundit every hour (sounds about right!).  In the first hour all 50 visitors would never have seen the link, and all would click.  In the next hour, on average only 25 of the 50 would be new to the link.  In the next hour, 12.5, and so on...

2. Instapundit readers tend to get distracted as they read down the page.  Suppose 100 new readers arrive at the site every hour. They all read the top link, but only 95% of them make it down to the second link. Of those remaining, 95% make it to the third link.  Now suppose that a link moves a fixed amount down the page every hour---say 10 slots.  Because 0.95^10=0.6 this would mean that in the first hour you&#039;d get 100 hits, in the second hour 60 hits, in the third hour 36...

About the time zones, you&#039;ll notice that this isn&#039;t &lt;i&gt;really&lt;/i&gt; exponential decay because the first couple hours don&#039;t decay fast enough.  This may be the result of the west coast waking up---essentially, the arrival rate of new visitors would be increasing in the first few hours, counteracting the decay.  Also, the 22nd hour might appear to be an outlier if it hadn&#039;t been completed yet when the photo was taken.</description>
		<content:encoded><![CDATA[<p>Here are two simplified models of the world that would lead to exponential decay:</p>
<p>1.  Instapundit readers arrive at the site randomly, and once there they read all the posts they have not read yet. As time goes on more and more of the readers are repeat visitors, and they don&#8217;t re-read.  To take one simple example, suppose there were 100 readers and each had a 50% chance of visiting Instapundit every hour (sounds about right!).  In the first hour all 50 visitors would never have seen the link, and all would click.  In the next hour, on average only 25 of the 50 would be new to the link.  In the next hour, 12.5, and so on&#8230;</p>
<p>2. Instapundit readers tend to get distracted as they read down the page.  Suppose 100 new readers arrive at the site every hour. They all read the top link, but only 95% of them make it down to the second link. Of those remaining, 95% make it to the third link.  Now suppose that a link moves a fixed amount down the page every hour&#8212;say 10 slots.  Because 0.95^10=0.6 this would mean that in the first hour you&#8217;d get 100 hits, in the second hour 60 hits, in the third hour 36&#8230;</p>
<p>About the time zones, you&#8217;ll notice that this isn&#8217;t <i>really</i> exponential decay because the first couple hours don&#8217;t decay fast enough.  This may be the result of the west coast waking up&#8212;essentially, the arrival rate of new visitors would be increasing in the first few hours, counteracting the decay.  Also, the 22nd hour might appear to be an outlier if it hadn&#8217;t been completed yet when the photo was taken.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Darrell</title>
		<link>http://neoneocon.com/2008/12/01/the-curve-of-an-instalanche/#comment-94833</link>
		<dc:creator>Darrell</dc:creator>
		<pubDate>Tue, 02 Dec 2008 05:03:53 +0000</pubDate>
		<guid isPermaLink="false">http://neoneocon.com/2008/12/01/the-curve-of-an-instalanche/#comment-94833</guid>
		<description>I think it is indicative of the number of people who just look at the first page of the site, the diminishing numbers are the ones that actually scroll down the page to see all of the new links posted since they checked last (like me). Blog dynamics?I saw the link and thought, I already read the best entry on that subject, then I checked the link and smiled, congrats Neo. Thats pretty close to the big time. Way to go.</description>
		<content:encoded><![CDATA[<p>I think it is indicative of the number of people who just look at the first page of the site, the diminishing numbers are the ones that actually scroll down the page to see all of the new links posted since they checked last (like me). Blog dynamics?I saw the link and thought, I already read the best entry on that subject, then I checked the link and smiled, congrats Neo. Thats pretty close to the big time. Way to go.</p>
]]></content:encoded>
	</item>
</channel>
</rss>

